Question

When four hydrogen nuclei combine to form a helium nucleus in the interior of sun, the loss in mass is 0.0265 a.m.u. How much energy is released?

Answer 1

Given that, Δm = 0.0265 a.m.u.

1 a.m.u. = 1.66 × 10^-27 kg

E = Δm × 1.66 × 10^-27 × C²J

E = ${\displaystyle \frac{0.0265\times 1.66\times {10}^{-27}\times {(3\times {10}^8)}^2}{1.6\times {10}^{-13}}\text{MeV}}$

E = ${\displaystyle \frac{265}{{10}^4}\times \frac{166\times {10}^{-27-2}\times 9\times {10}^{16}}{1.6\times {10}^{-13}}}$

E = ${\displaystyle \frac{265\times 166\times 9\times {10}^{-29+16-4}}{16\times {10}^{-14}}}$

E = ${\displaystyle \frac{2385\times 83}{8}\times {10}^{-13+14-4}}$

E = ${\displaystyle \frac{2385\times 83}{8\times 1000}}$

E = ${\displaystyle \frac{197955}{8\times 1000}}$

E = ${\displaystyle \frac{197.955}{8}}$

E = 24.7 MeV